Game Development Reference
In-Depth Information
(b) Applying Equations (13.32)-(13.35):
p
0
= b
0
= (3, 5)
v
0
= 3(b
1
−b
0
) = 3[(6, 1)−(3, 5)] = [9,−12]
v
1
= 3(b
3
−b
2
) = 3[(5, 5)−(0, 3)] = [15, 6]
p
1
= b
3
= (5, 5)
(c) Using Equation (13.19):
p(t) = b
0
+ t(3b
1
−3b
0
) + t
2
(3b
0
−6b
1
+ 3b
2
) + t
3
(−b
0
+ 3b
1
−3b
2
+ b
3
)
3
3
3
5
6
1
3
5
3
5
6
1
0
3
+ t
2
=
+ t
−3
−6
+ 3
−
3
5
5
5
6
1
0
3
+ t
3
+ 3
−3
+
3
5
−12
−27
18
2
−6
+ t
2
+ t
3
=
+ t
3
5
−12
−27
18
2
−6
+ 0.40
2
+ 0.40
3
(d) p(t) =
+ 0.40
3.00
5.00
3.56
2.70
3.60
−4.80
−4.32
2.88
1.28
−0.38
=
+
+
+
=
(e) Using Equation (13.5):
−27
18
−12
2
−6
v(t) = c
1
+ 2c
2
t + 3c
3
t
2
=
+ 3t
2
+ 2t
−54
36
−12
60
−18
+ t
2
=
+ t
−54
36
−12
60
−18
−12
−21.6
14.4
9.60
−2.88
−3.00
−0.48
+0.40
2
(f) v(0.40) =
+0.40
=
+
+
=
−12
−54
36
60
−18
9.00
−12.00
+ 0.00
2
v(0.00) =
+ 0.00
=
−12
−54
36
60
−18
15.00
6.00
+ 1.00
2
v(1.00) =
+ 1.00
=
4. 1 = B
0
(t) + B
1
(t) + B
2
(t)
= (1−t)
2
+ 2(1−t)t + t
2
= (1−2t + t
2
) + (2t−2t
2
) + t
2
= 1
5. All four control points should be in the same place.
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